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I have a question regarding capacitors and their charge neutrality. When capacitors are used in circuits, the assumption is often made that the plates of the capacitors have equal and opposite charges. I was wondering why this is the case.

I have done some research. One source, The Feynman Lectures on Physics (Vol. 2) explains (Ch. 22):

"We assume that the plates and the wires are perfect conductors. We also assume that the insulation between the plates is perfect, so that no charges can flow across the insulation from one plate to the other. Next, we assume that the two conductors are close to each other but far from all others, so that all field lines which leave one plate end up on the other. Then there are always equal and opposite charges on the two plates and the charges on the plates are much larger than the charges on the surfaces of the lead-in wires. Finally, we assume that there are no magnetic fields close to the capacitor."

I do not fully understand this argument. As a starting point, I don't understand why, if the plates are close to each other, all field lines which leave one plate end up on the other. I understand that field lines can terminate on negative charges, but can't they also just go off to infinity (thinking of a positive, point charge)? Moreover, even if all the field lines from one plate did terminate on the other, I do not see why this would imply the charges on the plates are equal and opposite. I am not sure if there is a mathematically rigorous argument for this, or if this is more of an intuitive argument.

A second argument I have seen involves the fact that batteries simply transport charge. These arguments typically take the example of a battery connected directly to the two ends of a capacitor. Assuming the system starts off charge neutral, it is clear that the two plates must have equal and opposite charges -- batteries do not create/destroy charge (of course) and remain charge neutral. I have found this argument in many places on this StackExchange -- one that I particularly like is found here. While I can appreciate this simple example, it does not seem sufficient to me.

Consider a more complex circuit, containing multiple capacitors. I have illustrated one such circuit below, but I'm sure one can imagine even more extreme cases (many capacitors, inductors, resistors, etc.).

More "complex" circuit

Now, things seem to get a little more complicated. Let's say the battery takes some charge from the bottom plate of C2 and transports it to the top plate of C1. Charge conservation is maintained, but the plates do not have equal and opposite charges. I can see one problem with this: the top plates of C2 and C1 now have different potentials, which would mean the system is not in steady state (current will flow through R1). However, I am hoping to find a justification for this that works beyond steady-state (one of my motivations for studying this is for high-frequency circuits).

This question interests me from an electrical engineering and circuits perspective. Often, when doing circuit analysis, any current that enters one of the capacitor's plates is assumed to exit the other plate. In other words, current is often envisioned traveling through the capacitor (despite the fact that no current actually flows between the two plates). Of course, this assumption is valid if the capacitor plates strictly maintain equal and opposite charges. I am just not sure why this must be true.

I have spent a while researching this and have not found concrete answers -- any help would be very much appreciated. If possible, I would really appreciate an answer that is mathematical (based on Maxwell's Equations, or other fundamental ideas). Thank you!

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  • $\begingroup$ Can you use gauss' law to prove that two plates have equal and opposite charges, assuming field lines starting from one plate end up on another $\endgroup$
    – HS Singh
    May 29 '20 at 3:49
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    $\begingroup$ "(despite the fact that no current actually flows between the two plates)" - that's not a true statement as written. It's true that (ideally) no electric charge flows between the plates of a capacitor. $\endgroup$ May 29 '20 at 3:59
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Often, when doing circuit analysis, any current that enters one of the capacitor's plates is assumed to exit the other plate.

We can assume this because when we inject an electron on one plate, the field it produces will repel other free charges around it. If the nearest free charges are on the other plate, then those are the ones that will get repelled, leading to the current out of one terminal being equal to the current in the other.

Of course you can also arrange, for example, for both plates to have some potential relative to your reference ground node. If a net charge moves in or out of the capacitor to change this potential, then you would model that with a parasitic capacitance between the two terminals of your capacitor and some other location in the circuit. This parasitic capacitance would account for electric field lines that go from the capacitor structure to "somewhere else" rather than originating on one plate and terminating on the other.

one of my motivations for studying this is for high-frequency circuits

In high frequency circuits you won't be assuming that a metal object is an equipotential. If you make your two "plates" larger than ~1/10 of the wavelength associated with the highest frequencies in your circuit, you will create a distributed structure rather than a lumped one. If the "plates" are very long and skinny, you have made a transmission line, for example. Then you will find that signals propagate along the structure as waves, with behavior dictated by the balance of the capacitance and inductance of the structure.

At some level you should also remember that all of our lumped circuit analysis is an approximation, based on certain simplifying assumptions about the nature of the circuit. If the lumped circuit model of a capacitor isn't adequate for explaining some particular circuit or device, you may have to perform a more detailed analysis, for example using Poisson's equation to analyze an electrostatic structure, or Maxwell's equations to analyze situations where magnetic and electric fields interact with the structure of the circuit (i.e. high-frequency situations).

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  • $\begingroup$ Nice. I remember taking into account these parasitic capacitors (and inductors) in an electromagnetic compatibility course I took (years ago). $\endgroup$ May 29 '20 at 4:07
  • $\begingroup$ Thank you for the response! Regarding your explanation of an electron entering one plate and then popping off the other, is there any rigorous, mathematical formulation for this? I can envision this conceptually, but it seems almost too simple (one comes in, one pops off, and then everything is fine). Perhaps I am overthinking it. $\endgroup$
    – dts
    May 29 '20 at 4:11
  • $\begingroup$ @dts "is there any rigorous, mathematical formulation for this?" - What specific quantity about this phenomenon do you want to be able to calculate? $\endgroup$ May 29 '20 at 4:18
  • $\begingroup$ @probably_someone in this particular example, I am hoping to calculate why adding one charge to one of the plates causes exactly one charge to pop off the other. Intuitively, I can see this as a possibility. But I can also see other possibilities, such as the charges rearranging in the other plate to "negate" the effect of adding that one charge (without any charge actually leaving the plate). Sorry, I know this sounds vague, I just cannot find any resources that give a rigorous treatment of this phenomenon, so I'm a little lost. $\endgroup$
    – dts
    May 29 '20 at 4:24
  • $\begingroup$ @dts If you have two sheets of unequal charge, then the fields outside those sheets will no longer exactly cancel. You only get zero field outside the capacitor if there are equal and opposite charges on the plates. So if you add an electron to the negative plate, then the field from that plate will increase, which means there will now be a net electric field behind the positive plate, which will push electrons away from the positive plate until the electric field behind the positive plate is zero again (which happens when there's one less electron on the positive plate). $\endgroup$ May 29 '20 at 4:33
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To cover the points not covered in the other answer:

As a starting point, I don't understand why, if the plates are close to each other, all field lines which leave one plate end up on the other. I understand that field lines can terminate on negative charges, but can't they also just go off to infinity (thinking of a positive, point charge)?

The usual assumption made when discussing ideal parallel-plate capacitors is that they are well approximated by infinite sheets of charge. Infinite sheets of charge emit a constant electric field perpendicular to the plates, regardless of distance from the plates. This means that, in the region between the plates, the field lines propagate perpendicular to both plates, extending from one to the other. In the region outside the capacitor, the two constant fields cancel exactly, so the field outside the ideal capacitor is zero.

Of course, real capacitors are made of plates with finite area, so there is always some fringing field that extends from one plate to another around the edges of the plates, but isn't quite perpendicular to the plates. In addition, the parallel-plate capacitor from far away looks like an electric dipole, so outside the capacitor there will also be a dipole field propagating through space. Typically we build capacitors taking these corrections into account. The relative importance of the fringing and dipole fields decreases when the area of the plates are increased, so generally we build capacitors with plates that are large enough that we can treat the capacitor as an ideal capacitor, neglecting the fringing and dipole fields without too much cost to accuracy.

Moreover, even if all the field lines from one plate did terminate on the other, I do not see why this would imply the charges on the plates are equal and opposite. I am not sure if there is a mathematically rigorous argument for this, or if this is more of an intuitive argument.

I'm not sure that Feynman is actually implying that in this passage. The word "then" in the passage refers to the impact of all of the assumptions taken up to that point, not just the last one. And, in fact, if you continue to read, Feynman declares exactly this:

"Since we imagined that the two plates are in some way isolated from the rest of the world, the total charge on the two plates must be zero; if there is a charge Q on the upper plate, there is an equal, opposite charge −Q on the lower plate."

You can imagine, I guess, a capacitor that starts with some kind of net charge on both of its plates, but this is not a component that would actually ever appear in circuit analysis. A net electric charge is a difficult thing to maintain in the real world - it's quickly diluted and/or neutralized by contact with other objects. So electrical components in circuits are generally assumed to at least start off electrically neutral.

Let's say the battery takes some charge from the bottom plate of C2 and transports it to the top plate of C1. Charge conservation is maintained, but the plates do not have equal and opposite charges.

The bottom plates of both capacitors are at exactly the same potential, since they're connected by a bare wire. If, at some instant, some charge left the bottom plate of C2 without being accompanied by some charge leaving the bottom plate of C1, then you would be left with two points connected by a bare wire that are at different potentials. This is a contradiction. If you tried to create such a condition, current would instantly flow to fix it (since the wire connecting them has zero resistance, instantaneous flow of charge is allowed). So the battery can only manipulate the charge of both capacitors at once.

This is true regardless of whether we're talking about steady-state current. After all, in an AC circuit of this construction, the currents through each capacitor are exactly in phase (since the only other component in the circuit, the resistor, does not alter the phase, only the magnitude of the current). As a result, the charge as a function of time on each capacitor changes exactly in sync, reaching maxima and minima at exactly the same time.

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  • $\begingroup$ "In addition, the capacitor from far away looks like an electric dipole, so outside the capacitor there will also be a dipole field propagating through space." - is this literally true? $\endgroup$ May 29 '20 at 4:22
  • $\begingroup$ @AlfredCentauri If you get far enough away, the field from a charged square of finite area looks like the field from a point charge. So if you get far enough away from a capacitor, the field will look like the field from two oppositely-charged point charges right next to each other: in other words, an electric dipole. $\endgroup$ May 29 '20 at 4:24
  • $\begingroup$ I appreciate your response, the points you make about Feynman's explanation clarifies things. Regarding the specific example circuit I gave, I can see what you mean with the bottom node having to maintain a constant potential. However, is there a way to generalize this idea? For instance, if I add a similar resistor between the bottom node of the two capacitors? Or if I replace the resistor with an inductor? I'm just curious if there is a foolproof explanation for this phenomenon, because (from my perspective) it seems like something we take for granted. $\endgroup$
    – dts
    May 29 '20 at 4:26
  • $\begingroup$ probably_someone, I'm thinking about cylindrical rolled foil capacitors right now, and my multi-pole expansion intuition is failing me. I'll think on it. $\endgroup$ May 29 '20 at 4:36
  • $\begingroup$ @dts A resistor won't do anything - they're not at the same potential anymore, but they are required to be separated by a particular potential difference. And current can still flow with infinite frequency across an ideal resistor, so this is still fixed instantly. An inductor is more interesting - it doesn't allow arbitrarily high frequencies of current, so it will induce some oscillation of charge. A resistor and an inductor will induce damped oscillation which eventually settles into an equilibrium configuration again. $\endgroup$ May 29 '20 at 4:40
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I will resort to a simple mechanical analogy here, which may make the physics easier to understand.

Imagine a mechanical system containing a coil spring. We suddenly apply a load to the system, which gets distributed among its various components, and it settles into the steady state after all the transients die away.

At that point we notice that the coil spring has been stretched out of its equilibrium position, and is not moving. We cleverly insert force gauges into both ends of the spring and notice that they read exactly the same force, except with opposite signs.

Since the spring is stationary, the force pulling one way on one end must be exactly balanced by an equal and opposite force pulling the other way on the other end.

By exactly the same reasoning, at steady state the voltage on the two leads of the capacitor must be identical, but opposite in sign.

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Of course it's possible to construct a capacitor with different charges on the the different plates. But the principle of superposition, which says that electric fields from different charge distributions add linearly, says that it's not interesting.

Let's suppose you go back to the chapter in your E&M book before capacitors, when you were applying nonzero net charge to things. "Consider a conducting sphere, with a net charge of 10 nC, find the electric field everywhere." That kind of problem. Ten nanocoulombs is about how much charge you can put on a piece of Scotch Tape by breaking the glue bond with the next layer down on the roll; my students used to do a lab where they would make the tape pieces repel each other and estimate the charge. You can transfer the charge from a static surface to a conductor by touching them together.

Now take your conducting sphere, saw it in half, and use an (insulating) clamp to squeeze a piece of paper or other insulator between the halves. Voilá! It's a capacitor with $C=\epsilon A/d$, where $\epsilon$ is the permittivity of your insulator, $A$ is the area of your cut, and $d$ is your insulator thickness. The smaller your $d$, the better the parallel-plate approximation is; the outer shape doesn't matter unless you're doing high-frequency stuff. Touch the two halves of your sphere-capacitor with the two terminals of a battery and you'll move charge $Q=CV$ from the one half to the other, but you can also put charge on either half just the way you did before you knew what capacitors were.

If the two halves of your sawed-in-half-sphere capacitor have charges $Q_\text{top}$ and $Q_\text{bottom}$, finding the field everywhere by directly integrating is hard. But, by the principle of superposition, the charges will have the same distribution as on a sphere with total charge $Q_\text{top}+Q_\text{bottom}$ superimposed on a parallel-plate capacitor with charges $\pm(Q_\text{top}-Q_\text{bottom})/2$ on the cut surfaces. This approximation gets better and better as the gap between the halves $d$ gets smaller; in the limit $d\to 0$, the parallel-plate capacitor doesn't have any fringing field at all.


In circuit analysis, why don't we care about the electric field emanating from the net-charged parts of a capacitor? We can actually quantify how much we care about this stray field, because we have a word for a circuit element where the energy is stored in an electric field over some volume of space. If the energy is stored in the electric field, the device is ... a capacitor. You only care about the stray fields when the stray capacitance is large.

Suppose that you have a circuit where some power supply is connected to a $1\rm\,\mu F$ capacitor and a meter of coaxial cable. The coaxial cable presents a parallel capacitance of perhaps $100\rm\,pF$, a tiny correction. The stray capacitance between one plate of the cable and one conductor in the cable? The parallel-plate formula $C=\epsilon A/d$ isn't right for that geometry, but the arguments that went into it are, and the distance $d$ between one plate of a capacitor and one conductor in the cable is huge, so that the corresponding stray capacitance is tiny. There's just no energy there.

We don't talk about the net charge on a capacitor because the energy stored in a capacitor doesn't come from the monopole term of the charge distribution; it comes from the dipole.

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This might help for your intuition

The assumption that you mentioned is only used for deriving the capacitance of a capacitor(using Gauss law). But in reality standard capacitors always have equal and opposite charges on it's plates while connected in a circuit (watch the video).

Why does the field confined only inside the capacitor?

It's indeed an application of Gauss law! I would like to give a hint, so that you could build other things on your own. Consider two infinitely large charged sheets(oppositely charged) kept at some distance parallel to each other. Now, what is the electric field inside and outside the charged plates?

Remember, in reality, capacitor plates are kept very close, so that a point inside the capacitor see the plates as (so called)infinitely large.

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Then there are always equal and opposite charges on the two plates

This not true if 2 or more capacitors are connected in series. For 2 capacitors in series, ideally the charge on the "inner" plates would be 0.

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